Question

(1-(a^-n+b^-n)/(a^-n-b^-n))^-2

Answer 1

(1 - (a^(-n) + b^(-n))/(a^(-n) - b^(-n)))^(-2) = ((a^(-n) - b^(-n) - a^(-n) - b^(-n))/(a^(-n) - b^(-n)))^(-2) = (-2•b^(-n)/(a^(-n) - b^(-n)))^(-2) = ((-2/b^n)/(1/a^n - 1/b^n))^(-2) = ((-2/b^n)/((b^n - a^n)/(a^n•b^n)))^(-2) = (-2/((b^n - a^n)/a^n))^(-2) = (-2•a^n/(b^n - a^n))^(-2) = (b^n - a^n))^2/(4•a^(2n))

Answer 2

To simplify the expression (1 - (a^(-n) + b^(-n))/(a^(-n) - b^(-n)))^(-2), we'll first simplify the fraction inside the parentheses and then raise it to the power of -2.

Step 1: Simplify the fraction inside the parentheses: Let's rewrite the fraction with a common denominator, which is (a^(-n) - b^(-n)):

(a^(-n) + b^(-n))/(a^(-n) - b^(-n))

To get the common denominator, we need to multiply the numerator and denominator by (a^(-n) + b^(-n)):

[(a^(-n) + b^(-n))/(a^(-n) - b^(-n))] * [(a^(-n) + b^(-n))/(a^(-n) + b^(-n))] = (a^(-n) + b^(-n))^2 / [(a^(-n))^2 - (b^(-n))^2]

Now, recall that (a^(-n))^2 = a^(-2n) and (b^(-n))^2 = b^(-2n):

= (a^(-n) + b^(-n))^2 / (a^(-2n) - b^(-2n))

Step 2: Raise the fraction to the power of -2:

[(a^(-n) + b^(-n))^2 / (a^(-2n) - b^(-2n))]^(-2)

To raise a fraction to a negative exponent, we swap the numerator and denominator and change the exponent's sign:

= [(a^(-2n) - b^(-2n)) / (a^(-n) + b^(-n))]^2

Therefore, the simplified expression is:

[(a^(-n) + b^(-n))/(a^(-n) - b^(-n))]^(-2) = [(a^(-2n) - b^(-2n)) / (a^(-n) + b^(-n))]^2